The Cantor ternary set is totally disconnected

I begin by assuming that by the $C_n$ you mean the usual closed sets whose intersection is the Cantor ternary set:

• $C_1 = [0,1]$;
• $C_2 = [0,\frac 13] \cup [ \frac 23 , 1 ]$;
• $C_3 = [0,\frac 19] \cup [\frac 29,\frac 13] \cup [\frac 23,\frac 79] \cup [\frac 89, 1]$;
• etc.

Note that for each $n$ the set $C_n$ is made up of disjoint closed intervals of length $3^{-(n-1)}$, and that these intervals are therefore separated from each other. Also note that if $x,y \in C_n$ are such that $3^{-(n-1)} < |x-y|$, then $x,y$ belong to different closed intervals making up $C_n$.

Given distinct $x,y \in C$ essentially by the Archimedean property there must be an $n$ such that $3^{-(n-1)} < |x-y|$, and as $x,y \in C_n$ it follows that they belong to different closed intervals making up $C_n$. Let $I$ be the closed interval in $C_n$ containing $x$. It follows that $x \in C \cap I$ and $y \in C \setminus I$, and these sets are separated.

Let $S$ be the set of open intervals that are removed from $[0,1]$ to produce the Cantor set $C.$ The members of S are pair-wise disjoint, and the sum of their lengths is $\frac {1}{3}\sum_{n=0}^{\infty}\left(\frac {2}{3}\right)^n=1.$

So no non-empty open interval $J\subset [0,1]$ is a subset of $C.$ Otherwise $T=\{J\}\cup S$ would be a set of pair-wise disjoint open intervals, each a subset of $[0,1],$ with the sum of the lengths of the members of $T$ being greater than $1.$

So when $x,y\in C$ with $x<y$ we have $(x,y)\not \subset C,$ so there exists $z\in (x,y)$ \ $C.$ So let $A=[0,z)\cap C$ and $B=(z,1]\cap C.$